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Chapter 0: Problem 117

Solve each radical equation. Check all proposed solutions. $$\sqrt{x+3}=x-3$$

Short Answer

Expert verified
The solution to the radical equation is \(x = 6\).

Step by step solution

01

Isolate the square root

The first step is to isolate the square root. In the given equation \(\sqrt{x+3}=x-3\), the square root is already isolated.
02

Squaring both sides

In order to remove the square root, you'll have to square both sides of the equation: \((\sqrt{x + 3})^2 = (x - 3)^2\). When you do that, you'll get \(x + 3 = x^2 - 6x + 9\).
03

Rearrange into quadratic equation

From the last step, we do have \(x + 3 = x^2 - 6x + 9\), if we move all terms into one side of the equation, we get \(x^2 - 7x + 6 = 0\).
04

Factorize quadratic equation

You can factorize the quadratic equation as follows \((x - 1)(x - 6) = 0\)
05

Solve for x

Now, set each factor equal to zero and solve: \(x - 1 = 0\) gives \(x = 1\), and \(x - 6 = 0\) gives \(x = 6\)
06

Check proposed solutions

By substituting x=1 back into the original equation, you get: \(\sqrt{1+3} = 1 - 3\), which simplifies to \(\sqrt{4} = -2\), which is not true. Following similar process for x=6, you get: \(\sqrt{6+3} = 6 - 3\), which simplifies to \(\sqrt{9} = 3\), which is true. The solution x=1 is an extraneous solution, we discard it.

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Most popular questions from this chapter

$$\text { Solve for } t: \quad s=-16 t^{2}+v_{0} t$$

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Perform the indicated operations. $$\frac{1}{x^{n}-1}-\frac{1}{x^{n}+1}-\frac{1}{x^{2 n}-1}$$

Perform the indicated operations. Simplify the result, if possible. $$\frac{a b}{a^{2}+a b+b^{2}}+\left(\frac{a c-a d-b c+b d}{a c-a d+b c-b d} \div \frac{a^{3}-b^{3}}{a^{3}+b^{3}}\right)$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I can check inequalities by substituting 0 for the variable: When 0 belongs to the solution set, I should obtain a true statement, and when 0 does not belong to the solution set. I should obtain a false statement.
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